Find the local maximum and minimum values and saddle point(s) of the function. I

Question

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list, If an answer does not exist, enter ONE.) f(x, y) = xy + 8/x + 8/y local maximum value(s) local minimum value(s) saddle point(s) (x, y, f) = Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list If an answer does not exist, enter ONE.) f(x, y) = 3y cos(x), 0 lessthanorequalto x lessthanorequalto 2 pi local maximum value(s) local minimum value(s) saddle point(s) (x, y, f) =

Explanation / Answer

f(x , y) = xy + (8/x) + (8/y)

fx = y - 8/x2 + 0

==> fx = y - (8/x2)

fy = x + 0 - 8/y2

==> fy = x - (8/y2)

fx = 0 ==> y = 8/x2

fy = 0 ==> x = (8/y2)

substitute y = 8/x2 in x = (8/y2)

==> x = 8/(8/x2)2

==> x = 8/(64/x4)

==> x = x4/8

==> x - x4/8 = 0

==> x(1 - x3/8) = 0

==> x = 0 , x = 2

x cannot be zero since y becomes infinity

when x = 2 , y = 8/(2)2 = 2

Hence critical point is (2 , 2)

fxx = 0 - 8(-2)/x3 = 16/x3

fxx at (2 , 2) = 16/8 = 2 > 0

fyy = 0 - 8(-2)/y3 = 16/y3

fyy at (2 , 2) ==> 16/8 = 2

fxy = 1 - 0 = 1

D = fxxfyy - (fxy)2

==> D = (2)(2) - (1)2 = 4 - 1 = 3 > 0

as D > 0 and fxx > 0 , the function has relative minimum at (2 , 2)

minimum value = 2(2) + (8/2) + (8/2) = 4 + 4 + 4 = 12

maximum value = DNE

saddle points = DNE

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